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Given 3 collinear points A, B,C. Find D (by constructing) if R(ABCD)=1.7/3 (projective geometry). Can someone help me with this problem.

user571977
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1 Answers1

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You can use the construction described here to construct a projective scale, i.e. a set of points which are equidistant with respect to a chosen point of infinity. First pick $A$ as the point at infinity, $B$ as $0$ and $C$ as $1$. Then construct $17$ steps, so you reach a point $P$ which satisfies $(A,B;C,P)=17$. Then go for the denominator. Choose $B$ as infinity, $A$ as $0$ and $P$ as $1$ and construct 30 steps for a point $D$ with $(B,A;P,D)=30$ and $(A,B;C,D)=\frac{17}{30}$.

This is easily said and described, but when you actually try doing this with a real straightedge construction, you will find elements bunched together too closely to distinguish, so it's not a terrbily practical aproach. What's your use case? What tools do you have at hand? Do you have any geometrical representation of the desired ratio, or just numbers? Does it need to be exact?

MvG
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